If `lim_(xto0)(log_(e)cot((pi)/4-K_(1)x))/(tanK_(2)x)=1`, then

Options

A. `K_(1)=K_(2)`
B. `2K_(1)=K_(2)`
C. `K_(1)=2K_(2)`
D. `K_(1)=4K_(2)`

Answer

To solve the limit problem given by:

\[ \lim_{x \to 0} \frac{\log(\cot(\frac{\pi}{4} - k_1 x))}{\tan(k_2 x)} = 1 \]

we will follow these steps:

### Step 1: Analyze the limit
As \( x \to 0 \), both the numerator and denominator approach 0, leading to the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and denominator:

1. **Differentiate the numerator**:
\[ \frac{d}{dx} \left( \log(\cot(\frac{\pi}{4} - k_1 x)) \right) = \frac{-k_1 \csc^2(\frac{\pi}{4} - k_1 x)}{\cot(\frac{\pi}{4} - k_1 x)} \]

2. **Differentiate the denominator**:
\[ \frac{d}{dx} (\tan(k_2 x)) = k_2 \sec^2(k_2 x) \]

### Step 3: Rewrite the limit
Now we rewrite the limit using the derivatives obtained:

\[ \lim_{x \to 0} \frac{-k_1 \csc^2(\frac{\pi}{4} - k_1 x)}{k_2 \sec^2(k_2 x)} \]

### Step 4: Evaluate the limit as \( x \to 0 \)
As \( x \to 0 \):
- \( \cot(\frac{\pi}{4}) = 1 \) implies \( \csc^2(\frac{\pi}{4}) = 2 \).
- \( \sec^2(0) = 1 \).

Thus, the limit simplifies to:

\[ \lim_{x \to 0} \frac{-k_1 \cdot 2}{k_2 \cdot 1} = \frac{-2k_1}{k_2} \]

### Step 5: Set the limit equal to 1
According to the problem statement, we set this equal to 1:

\[ \frac{-2k_1}{k_2} = 1 \]

### Step 6: Solve for \( k_2 \)
Rearranging gives:

\[ -2k_1 = k_2 \]

Thus, we can express \( k_2 \) in terms of \( k_1 \):

\[ k_2 = -2k_1 \]

### Final Result
The relationship between \( k_1 \) and \( k_2 \) is:

\[ k_2 = -2k_1 \]